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Refrigerant 134a (R134a) enters the coils of the evaporator of a refrigeration system as a saturated liquid-vapour mixture at a pressure of 160 kPa. The refrigerant absorbs 180 kJ of heat from the cooled space, which is maintained at -5oC, and leaves as saturated vapour at the same pressure. (a) What is the temperature of the R134a liquid-vapour mixture as it enters the system? (b) What is the specific enthalpy of the R134a liquid-vapour mixture as it exits the system? (c) What is the entropy change of the R134a? (d) What is the total entropy change for this process?

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danialamin

Answer:

Part a: The temperature of the mixture at 160kPA is 257.68K.

Part b:The specific enthalpy of the mixture as it exits the system is 237.97 kJ/kg.

Part c:The entropy change for the mixture  is 0.6999kJ/K

Part d: The total entropy change is 0.027 kJ/K

Explanation:

Part a

The temperature is as viewed from the the table of Saturated properties for the Refrigerant 134a for Pressure of 160kPa.

[tex]T_{sat\rightarrow 160kPa}=-15.62^o C\\T_{sat\rightarrow 160kPa}=-15.62+273K\\T_{sat\rightarrow 160kPa}=-15.62^o C\,\, or \,\, 257.28 K[/tex]

The temperature of the mixture at 160kPA is 257.68K

Part b

For specific enthalpy from the same table of Saturated properties for the Refrigerant 134a for Pressure of 160kPa for the vapour phase which is given as

[tex]h_g=237.97 kJ/kg[/tex]

The specific enthalpy of the mixture as it exits the system is 237.97 kJ/kg.

Part c

The entropy change for the mixture is given as

[tex]s=\frac{\Delta Q}{T}[/tex]

Here

ΔQ is the heat in which is given as 180kJ

T is the temperature of the mixture which is calculated in part a as 257.28K

so the equation becomes.

[tex]s_1=\frac{Q_{in}}{T}\\s_1=\frac{180 kJ}{257.28K}\\s_1=0.6999 kJ/K[/tex]

The entropy change for the mixture  is 0.6999kJ/K

Part d

For overall entropy change, the entropy change in the cooled space is also to be calculated such that

Q is the heat in which is given as 180kJ

T is the temperature of the cooled space which is given as -5°C or 268K

so the equation becomes.

[tex]s_2=-\frac{ Qout}{T}\\s_2=-\frac{180 kJ}{268K}\\s_2=-0.672 kJ/K[/tex]

So the total entropy change is

[tex]\Delta s=\Delta s_1+\Delta s_2\\\Delta s_{gen} =0.699-0.672\\\Delta s_{gen} =0.027 kJ/K[/tex]

The total entropy change is 0.027 kJ/K

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